Continuity of the following function $f(x)=\sum_{k=1}^\infty\frac{x^kk!}{(x+1)(2x+1)\dots(kx+1)}$ $$f(x)=\sum_{k=1}^\infty\frac{x^kk!}{(x+1)(2x+1)\dots(kx+1)}$$
I'm asked to investigate this series and determine where it's continuous, but I have no idea where to even start with this.
I was able to determine that the series converges when 0<x<1, and diverges when x$\ge$1, using the raabe test, but after this I don't know how to to show continuity anywhere.
EDIT: I determined the convergence as follows:
$$\rho_k\equiv k\left(\frac{x^kk!}{(x+1)(2x+1)\dots(kx+1)}\frac{(x+1)\dots(kx+1)((k+1)x+1)}{x^{k+1}(k+1)!}-1\right)$$
$$\rho_k=k\left(\frac{(k+1)x+1}{(k+1)x}-1\right)=\frac{k}{(k+1)x}$$
$$\lim_{k\to\infty}\rho_k=\frac{1}{x}\lim_{k\to\infty}\frac{k}{k+1}=\frac{1}{x}$$
So the series converges when $\frac{1}{x}>1$ and diverges when $\frac{1}{x}<1$, so $x<1$ and $x>1$. When $x<0$ we have $\frac{1}{x}>1\Rightarrow 1<x$ which is a contradiction, so $x\nless0$
When $x=1$ we have
$$f(1)=\sum_{k=1}^\infty\frac{1^kk!}{(1+1)(2\cdot 1+1)\dots(k\cdot 1+1)}=\sum_{k=1}^\infty\frac{k!}{(k+1)!}=\sum_{k=1}^\infty\frac{1}{k+1}$$ which diverges
So we have $x\ge 1$ diverges and $0\le x<1$ converges
 A: As already established, the given series $$f(x)=\sum_{k=1}^\infty f_k(x),\qquad f_k(x)=\frac{x^k\ k!}{(x+1)\cdots(kx+1)}$$ converges if and only if $0\leqslant x<1$.
Moreover, it converges uniformly on $x\in[0,a]$ for any $0<a<1$ (by Weierstrass M-test, since $0\leqslant f_k(x)\leqslant f_k(a)$ for $0\leqslant x\leqslant a$), hence $f(x)$ is continuous (as each $f_k$ is).
It is not asked for, but the equality $\color{blue}{f(x)=x/(1-x)}$ follows from $$\sum_{k=1}^n f_k(x)=\frac{x}{1-x}\big(1-(n+1)f_n(x)\big),$$ which is easy to show using induction on $n$.
A: Manipulating the numerator and denominator of the sum slightly, it turns out that the infinite series can be expressed in terms of the hypergeometric function
$$f(x)=\sum_{k=0}^\infty\frac{(k+1)!}{(1+1/x)(2+1/x)\dots(k+1+1/x)}\\=\frac{x}{1+x}\sum_{k=0}^\infty\frac{(1)_k(2)_k}{(2+1/x)_k}\\=\frac{x}{1+x}{}_2F_1(1,2,2+1/x;1)$$
From considerations contained in this question, it can be shown that ${}_2F_1(1,2,2+1/x,z)$ is uniformly continuous as long as $|z|<1$. This leaves the present problem as an edge case, which has to be treated separately, due to the singularity present at $x=z=1$.
To assess continuity, we look at the integral representation
$$B(b,c-b){}_2F_1(a,b,c;1)=\int_0^1 x^{b-1}(1-x)^{c-b-a-1}$$
which from the definition of the beta integral gives,
$$f(x)=\frac{x}{1+x}\frac{B(2,1/x-1)}{B(2,1/x)}=\frac{x}{1-x}~~,~~ |x|<1$$
as @metamorphy pointed out in the comments. This is obviously uniformly continuous in any subset of the unit disk (not including its boundary).
